An infinite set is always "maintained"

[William]

On Thursday, September 1, 2022 at 10:09:42 AM UTC-3, WM wrote:
>
> If you argue that all numbers leave the intersection but [a set of] infinitely many numbers do not leave the contributing endsegments, then you are outside of logic and reason.

Indeed. But there is no unchanging set of infinitely many numbers that does not "leave the endsegments" . There is only a decreasing sequence of sets
with no last element.

[mitchr...@gmail.com]

How do you contain infinite set elements?

Mitchell Raemsch

[FromTheRafters]

mitchr...@gmail.com explained :

With curly brackets.

[Sergio]

....there are no maintenance people to "maintain" your infinite set.

Please use the curly brackets instead => { }

and electrify them to keep all infinity of the Ants inside.

[mitchr...@gmail.com]

That can't be done. Just as there are no countable infinities.

[WM]

William schrieb am Donnerstag, 1. September 2022 um 18:24:43 UTC+2:
> On Thursday, September 1, 2022 at 10:09:42 AM UTC-3, WM wrote:
> >
> > If you argue that all numbers leave the intersection but [a set of] infinitely many numbers do not leave the contributing endsegments, then you are outside of logic and reason.
>
> Indeed. But there is no unchanging set of infinitely many numbers that does not "leave the endsegments" .

There is an infinite set not leaving. Inclusion monotony proves that thre is an unchanging infinite subset to all the infinite sets.

>There is only a decreasing sequence of sets
> with no last element.

You cannot imagine a decreasing sequence of infinite sets?
Use the intervals a_n = [0, ω + 1/n].Their intersection ist infinite.

Regards, WM

[William]

On Thursday, September 1, 2022 at 5:35:28 PM UTC-3, WM wrote:
deed. But there is no unchanging set of infinitely many numbers that does not "leave the endsegments" .
> There is an infinite set not leaving. Inclusion monotony proves that thre is an unchanging infinite subset to all the infinite sets.

Nope. Not unchanging

> >There is only a decreasing sequence of sets
> > with no last element.
> You cannot imagine a decreasing sequence of infinite sets?

Sure I can. I can even imagine a decreasing sequence of infinite sets of natural numbers.

> Use the intervals a_n = [0, ω + 1/n].Their intersection ist infinite.

So what? The a_n are not sets of natural numbers
ω + 1/n is not a natural number. so [0, ω + 1/n] is not an interval of natural numbers.
b_n=[n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n). The intersection is empty
For each n, b_n is infinite. The *elements* of b_n are finite.

[Fritz Feldhase]

On Thursday, September 1, 2022 at 10:35:28 PM UTC+2, WM wrote:
>
> Inclusion monotony proves that thre is an unchanging infinite subset to all the infinite sets.

No, it doesn't, Du psychotisches Arschloch voller Scheiße!

> the intervals a_n = [0, ω + 1/n].

Mückenheim, Sie sind eindeutig nicht mehr ganz dicht.

[Sergio]

On 9/1/2022 3:35 PM, WM wrote:
> William schrieb am Donnerstag, 1. September 2022 um 18:24:43 UTC+2:
>> On Thursday, September 1, 2022 at 10:09:42 AM UTC-3, WM wrote:
>>>
>>> If you argue that all numbers leave the intersection but [a set of] infinitely many numbers do not leave the contributing endsegments, then you are outside of logic and reason.
>>
>> Indeed. But there is no unchanging set of infinitely many numbers that does not "leave the endsegments" .
>
> There is an infinite set not leaving.

Wrong, sets do not come and go as they please.

> Inclusion monotony proves that thre is an unchanging infinite subset to all the infinite sets.

And just how does it prove it ? what is it ?

it is: one unchanging infinite subset to all the infinite sets

so Prove it. I dont see how you can do that, "All infinite sets" ?

>
>> There is only a decreasing sequence of sets
>> with no last element.
>
> You cannot imagine a decreasing sequence of infinite sets?
> Use the intervals a_n = [0, ω + 1/n].Their intersection ist infinite.

that's redundant, (you can throw 1/n out)

a_n = a_n+1 = a_n+2 so what is the intersection ?



>
> Regards, WM

[WM]

William schrieb am Donnerstag, 1. September 2022 um 22:49:27 UTC+2:
> On Thursday, September 1, 2022 at 5:35:28 PM UTC-3, WM wrote:
> deed. But there is no unchanging set of infinitely many numbers that does not "leave the endsegments" .
> > There is an infinite set not leaving. Inclusion monotony proves that thre is an unchanging infinite subset to all the infinite sets.
> Nope. Not unchanging

Decreasing., But the infinite core is same in all infinite sets.

> > >There is only a decreasing sequence of sets
> > > with no last element.
> > You cannot imagine a decreasing sequence of infinite sets?
> Sure I can. I can even imagine a decreasing sequence of infinite sets of natural numbers.
> > Use the intervals a_n = [0, ω + 1/n].Their intersection ist infinite.
> So what? The a_n are not sets of natural numbers
> ω + 1/n is not a natural number. so [0, ω + 1/n] is not an interval of natural numbers.

But this example shows that you general refusal is unfounded.

> b_n=[n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n). The intersection is empty

No, the intersection is infinite since all sets contribute an infinite subset.

Regards, WM

[Sergio]

On 9/2/2022 7:38 AM, WM wrote:
> William schrieb am Donnerstag, 1. September 2022 um 22:49:27 UTC+2:
>> On Thursday, September 1, 2022 at 5:35:28 PM UTC-3, WM wrote:
>> deed. But there is no unchanging set of infinitely many numbers that does not "leave the endsegments" .
>>> There is an infinite set not leaving. Inclusion monotony proves that thre is an unchanging infinite subset to all the infinite sets.
>> Nope. Not unchanging
>
> Decreasing., But the infinite core is same in all infinite sets.

Still wrong, all endsegments are infinite, and the same size. there is no "core".

Core Ants.

>
>>>> There is only a decreasing sequence of sets
>>>> with no last element.
>>> You cannot imagine a decreasing sequence of infinite sets?
>> Sure I can. I can even imagine a decreasing sequence of infinite sets of natural numbers.
>>> Use the intervals a_n = [0, ω + 1/n].Their intersection ist infinite.
>> So what? The a_n are not sets of natural numbers
>> ω + 1/n is not a natural number. so [0, ω + 1/n] is not an interval of natural numbers.
>
> But this example shows that you general refusal is unfounded.

no, your example is just flat wrong, and bad, sloppy math, and is REJECTED by this newsgroup.

>
>> b_n=[n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n). The intersection is empty
>

> No, the intersection is bla bla...

Bla Bla Ants of the Intersection

>
> Regards, WM

[William]

All sets contribute an infinite subset, but not the same infinite subset. There is no common subset.

[Fritz Feldhase]

On Friday, September 2, 2022 at 2:39:00 PM UTC+2, WM wrote:
> William schrieb am Donnerstag, 1. September 2022 um 22:49:27 UTC+2:
> >

> > b_n = [n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n).

Well, let's say IN n [n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n).

> The intersection is empty

Right.

> No,

Yes.

> the intersection is infinite

No, it isn't.

> since all sets contribute an infinite subset.

Nope, this fact doe NOT imply that the intersection of all b_n (with n e IN) is infinite . Non sequitur.

Und jetzt hau endlich ab, Du Spinner!

[Ross A. Finlayson]

Oh, ..., I thought ants were mathematical.

Intersection of what?

[WM]

You are wrong. Show me the first b_n which has not infinitely many elements in common with all its predecessors. You can't.

But if you could, then, as soon as there is no longer an infinite common subset, the next state of affairs is a finite common subset by the very definition of endsegments, i.e., simple mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

To deny it by the analogy that
∀k ∈ ℕ ∃m ∈ ℕ: k < m ==> ∃m ∈ ℕ ∀k ∈ ℕ: k < m
is an invalid conclusion, is an invalid conclusion and shows naivety.

Regards, WM

[WM]

Fritz Feldhase schrieb am Freitag, 2. September 2022 um 16:42:51 UTC+2:
> On Friday, September 2, 2022 at 2:39:00 PM UTC+2, WM wrote:
> > William schrieb am Donnerstag, 1. September 2022 um 22:49:27 UTC+2:
> > >
> > > b_n = [n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n).
>
> Well, let's say IN n [n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n).
>
> > The intersection is empty
>
> Right.
>
> > No,
>
> Yes.
>
> > the intersection is infinite
>
> No, it isn't.

As soon as there is no longer an infinite common set, the next state of affairs is a finite set by the very definition of endsegments, i.e., simple mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

To overlook this shows incompetence in mathematics and logic.

[William]

On Friday, September 2, 2022 at 5:12:24 PM UTC-3, WM wrote:
> William schrieb am Freitag, 2. September 2022 um 16:34:04 UTC+2:
> > On Friday, September 2, 2022 at 9:39:00 AM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 1. September 2022 um 22:49:27 UTC+2:
> >
> > > > b_n=[n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n). The intersection is empty
> > > No, the intersection is infinite since all sets contribute an infinite subset.
> > All sets contribute an infinite subset, but not the same infinite subset. There is no common subset.
> You are wrong. Show me the first b_n which has not infinitely many elements in common with all its predecessors. You can't.

So what?

>
> But if you could, then, as soon as there is no longer an infinite common subset,

This never happens. So what? At every step there is a common subset, but it changes. There is no one set contained in all common subsets. There is no "infinite core".

--
William Hughes

[Sergio]

On 9/2/2022 3:13 PM, WM wrote:
> Fritz Feldhase schrieb am Freitag, 2. September 2022 um 16:42:51 UTC+2:
>> On Friday, September 2, 2022 at 2:39:00 PM UTC+2, WM wrote:
>>> William schrieb am Donnerstag, 1. September 2022 um 22:49:27 UTC+2:
>>>>
>>>> b_n = [n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n).
>>
>> Well, let's say IN n [n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n).
>>
>>> The intersection is empty
>>
>> Right.
>>
>>> No,
>>
>> Yes.
>>
>>> the intersection is infinite
>>
>> No, it isn't.
>
> As soon as there is no longer an infinite common set,

there is no common infinite set.


Go back to your potentially infinite book, and find none of the math there applies to infinite sets.

> the next state of affairs is a finite set by the very definition of endsegments, i.e., simple mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

that equation does not define an endsegment.

>
> To overlook this shows incompetence in mathematics and logic.

which you have in spades.

>
> To deny it by the analogy that
> ∀k ∈ ℕ ∃m ∈ ℕ: k < m ==> ∃m ∈ ℕ ∀k ∈ ℕ: k < m
> is an invalid conclusion, is an invalid conclusion and shows naivety.

red herring

>
> Regards, WM

[Ross A. Finlayson]

No it's:

∀k ∈ ℕ ∃m ∈ ℕ: k < m ==> ∀m2 ∈ ℕ ∃k2 ∈ ℕ ∃m3 ∈ ℕ: k2 < m3, ∃j ∈ ℕ: k2 < m, ~∃l ∈ ℕ, l < j, ~(m2< m3 ^ ~ m2 < m3)


I.e. there is "always only proving the properties of natural numbers
that have exists m>n while also not for zero".

I.e., saying "exists m , ..., in N", is follows "exists N, exists < in N, exists m in N, ...".

[Fritz Feldhase]

On Saturday, September 3, 2022 at 4:17:37 AM UTC+2, Sergio wrote:

> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>
> that equation does not define an endsegment.

Right. But together with

E(0) it defined the sequence (E_n)_(n e IN) /of endsegments/ recursively.

From that sequence we may get the set of all endsegments:

ENDSEG := {E_n c IN : n e IN}.

Then we may define:

endsegement(E) :<-> E e ENDSEG.

Of course we may also define /endsegment/ directly:

endsegement(E) :<-> En e IN: E = E_n ,

and then define the set ENDSEG of all endsegments using /endsegement/:

ENDSEG := {E c IN : endsegement(E)}.

> > To deny it by the analogy that
> > ∀k ∈ ℕ ∃m ∈ ℕ: k < m ==> ∃m ∈ ℕ ∀k ∈ ℕ: k < m

> > is an invalid conclusion, is an invalid conclusion and <bla bla>

Well, a "quantifier shift" is indeed a logical fallacy.

WM just likes to commit this error _whenever it's possible_.

> red herring

This idiot does not understand ANY math.

[WM]

William schrieb am Freitag, 2. September 2022 um 22:59:23 UTC+2:
> On Friday, September 2, 2022 at 5:12:24 PM UTC-3, WM wrote:
> > William schrieb am Freitag, 2. September 2022 um 16:34:04 UTC+2:
> > > On Friday, September 2, 2022 at 9:39:00 AM UTC-3, WM wrote:
> > > > William schrieb am Donnerstag, 1. September 2022 um 22:49:27 UTC+2:
> > >
> > > > > b_n=[n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n). The intersection is empty
> > > > No, the intersection is infinite since all sets contribute an infinite subset.
> > > All sets contribute an infinite subset, but not the same infinite subset. There is no common subset.
> > You are wrong. Show me the first b_n which has not infinitely many elements in common with all its predecessors. You can't.
> So what?

No take the consequences.

> >
> > But if you could, then, as soon as there is no longer an infinite common subset,
> This never happens. So what? At every step there is a common subset, but it changes. There is no one set contained in all common subsets.

As soon as there is no longer an infinite common set, the next state of affairs is a finite set by the very definition of endsegments, i.e., simple mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

To deny it by the analogy that
∀k ∈ ℕ ∃m ∈ ℕ: k < m ==> ∃m ∈ ℕ ∀k ∈ ℕ: k < m
is an invalid conclusion, is an invalid conclusion and shows naivety.

> There is no "infinite core".

Then the core must be reduced step by step. It is impossible to jump over more than one number lost.

Regards, WM

[Sergio]

WM has the super Ego of a PHD in physics, and thinks he knows Math exceptionally well, however in this newsgroup he cannot get a passing grade in
beginning algebra.

Makes me think of one of the math Books he wrote, I think most of it is plagerized now, the book covers too much ground, too fast, and many sections
seem chopped out of other books.

[William]

On Saturday, September 3, 2022 at 5:12:27 PM UTC-3, WM wrote:
> William schrieb am Freitag, 2. September 2022 um 22:59:23 UTC+2:
> > On Friday, September 2, 2022 at 5:12:24 PM UTC-3, WM wrote:

> > > But if you could, then, as soon as there is no longer an infinite common subset,
> > This never happens. So what? At every step there is a common subset, but it changes. There is no one set contained in all common subsets.
> As soon as there is no longer an infinite common set,

This does not happen.

[Tom Bola]

Sergio schrieb:

It is also an abhoring kind of person...

[FromTheRafters]

WM formulated the question :

> William schrieb am Freitag, 2. September 2022 um 22:59:23 UTC+2:
>> On Friday, September 2, 2022 at 5:12:24 PM UTC-3, WM wrote:
>>> William schrieb am Freitag, 2. September 2022 um 16:34:04 UTC+2:
>>>> On Friday, September 2, 2022 at 9:39:00 AM UTC-3, WM wrote:
>>>>> William schrieb am Donnerstag, 1. September 2022 um 22:49:27 UTC+2:
>>>>>> b_n=[n,ω) is an interval of natural numbers. (ω is not an element of b_n
>>>>>> for any n). The intersection is empty
>>>>> No, the intersection is infinite since all sets contribute an infinite
>>>>> subset.
>>>> All sets contribute an infinite subset, but not the same infinite subset.
>>>> There is no common subset.
>>> You are wrong. Show me the first b_n which has not infinitely many elements
>>> in common with all its predecessors. You can't.
>> So what?
>
> No take the consequences.
>>>
>>> But if you could, then, as soon as there is no longer an infinite common
>>> subset,
>> This never happens. So what? At every step there is a common subset, but it
>> changes. There is no one set contained in all common subsets.
>
> As soon as there is no longer an infinite common set, the next state of
> affairs is a finite set by the very definition of endsegments, i.e., simple
> mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

If E(some natural number) is your endsegment, and it appears on both
sides of an equation, how does this equation 'define' what an
endsegment is?

[Tom Bola]

FromTheRafters schrieb:

WM can't hear you - too complicated given his IQ and he is also dreaming...

[Fritz Feldhase]

On Sunday, September 4, 2022 at 12:39:19 AM UTC+2, FromTheRafters wrote:
> WM formulated:

> >
> > by the very definition of endsegments, i.e., simple
> > mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> >

> [...] how does this equation 'define' what an
> endsegment is?

It's part of a recursive definition, you silly dimwit!

Of course, in addition we need E(1) = IN.

[Fritz Feldhase]

On Saturday, September 3, 2022 at 10:12:27 PM UTC+2, WM wrote:
>
> As soon as there is no longer an infinite common set, the next state of affairs is <bla>

Hint: For each and every n e IN there is a common infinte subset of the sets E(1), ..., E(n), namely E(n).

So no "next state of affairs".

You are incredible dumb, Mückenheim!

[FromTheRafters]

Fritz Feldhase wrote :

Thanks for part of a good answer.

[Fritz Feldhase]

You are welcome!

Funny thing: These "defining equations" allow for a simple proof (by induction) that for all n e IN: E(n) Is infinite (contrary to WM's claims).

Proof (by induction): E(1) = IN, hence E(1) is infinite. If E(n0) is infinite (for some n0 e IN), then certainly E(n0+1) = E(n0) \ {n0} is infinite (too). Hence (by induction) for all n e IN: E(n) is infinite. qed

It's a pitty that WM is not able to USE his

> > > > simple mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k}

for any real math.

[Sergio]

No worries, as equation is a set of characters, you can plow it back in, like fertilizer, thusly;

( E ∀ { k ∈ \ ℕ E = k 1 ) k ) k : } + (

or

} k { \ ) k ( E = ) 1 + k ( E : ℕ ∈ k ∀

[FromTheRafters]

Fritz Feldhase explained :

It shows the relation between one 'endsegment' and the next. He already
defined his endsegment as natural numbers greater than k when he
defined his FISONs (which are always finite, so no need for the "F") as
natural numbers less than or equal to k.

E(1) does not equal the naturals in muckymath. Initial segments are all
non-zero natural numbers less than or equal to k, and k+1 and beyond
are the successive endsegment indices.

Unless he switches to the naturals the rest of us use, in which case he
should 'define' the terms he is using.

[Fritz Feldhase]

On Sunday, September 4, 2022 at 3:25:38 PM UTC+2, FromTheRafters wrote:
> Fritz Feldhase explained :
> > On Sunday, September 4, 2022 at 1:46:21 PM UTC+2, FromTheRafters wrote:
> >> Fritz Feldhase wrote :
> >>> On Sunday, September 4, 2022 at 12:39:19 AM UTC+2, FromTheRafters wrote:
> >>>> WM formulated:
> >>>>>
> >>>>> by the very definition of endsegments, i.e., simple
> >>>>> mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> >>>>>
> >>>> [...] how does this equation 'define' what an
> >>>> endsegment is?
> >>>>
> >>> It's part of a recursive definition, you silly dimwit!
> >>>
> >>> Of course, in addition we need E(1) = IN.
> >>>
> >> Thanks for part of a good answer.
> >
> > You are welcome!
> >
> > Funny thing: These "defining equations" allow for a simple proof (by
> > induction) that for all n e IN: E(n) Is infinite (contrary to WM's claims).
> >
> > Proof (by induction): E(1) = IN, hence E(1) is infinite. If E(n0) is infinite
> > (for some n0 e IN), then certainly E(n0+1) = E(n0) \ {n0} is infinite (too).
> > Hence (by induction) for all n e IN: E(n) is infinite. qed
> >
> > It's a pitty that WM is not able to USE his
> >
> >>>>> simple mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> >
> > for any real math.
> >
> It shows the relation between one 'endsegment' and the next.

Indeed!

> He already defined his endsegment as <bla>

Nope.

Actually, E(k) = {n e IN : n >= k} for all k e IN
in the context of mückenmath.

Yes, I KNOW that this means that E(k) =/= IN \ F(k),
with F(k) = {n e IN : n <= k} for all k e IN.

> E(1) does not equal the naturals in muckymath.

It does, dumbo.

My advice: Shut up and fuck off!

Alternatively you might try to learn from your betters.

[WM]

As long as there is an infinite common set, the intersection is this infinite common set.

Regards, WM

[WM]

FromTheRafters schrieb am Sonntag, 4. September 2022 um 00:39:19 UTC+2:
> WM formulated the question :

> > As soon as there is no longer an infinite common set, the next state of
> > affairs is a finite set by the very definition of endsegments, i.e., simple
> > mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> If E(some natural number) is your endsegment, and it appears on both
> sides of an equation, how does this equation 'define' what an
> endsegment is?

With E(1) = ℕ all is said. But that should be known meanwhile.

Regards, WM

[WM]

Fritz Feldhase schrieb am Sonntag, 4. September 2022 um 04:21:25 UTC+2:
> On Saturday, September 3, 2022 at 10:12:27 PM UTC+2, WM wrote:
> >
> > As soon as there is no longer an infinite common set, the next state of affairs is <bla>
>
> Hint: For each and every n e IN there is a common infinte subset of the sets E(1), ..., E(n), namely E(n).

And other endsegments than indexed by each and every n e IN are not available and do not fill an infinite set and do not change the infinity of the common subset.

Regards, WM

[Sergio]

dangerous statement => "in the context of mückenmath" it causes finite bathtubs to potentially leak to infinity

>
> Yes, I KNOW that this means that E(k) =/= IN \ F(k),
> with F(k) = {n e IN : n <= k} for all k e IN.
>
>> E(1) does not equal the naturals in muckymath.

dangerous statement => "in muckymath" it causes Achilles to get Framed and lose big time to fat slow but rich turtle

>
> It does, dumbo.
>
> My advice: Shut up and fuck off!
>
> Alternatively you might try to learn from your betters.

I'm here, what is your question I can learn you?

[Sergio]

there isn't, so there is not.

>
> Regards, WM

[Sergio]

∀k ∈ ℕ: E(k+1) = E(k) \ {k} only shows the relationship between two adjacent endsegments.

it is not a step by step equation, that is bad math.

[Sergio]

why are they "not available" ?

What is your common subset ?


any answers or are you just making stuff up ?

[Gus Gassmann]

On Sunday, 4 September 2022 at 11:08:34 UTC-3, WM wrote:
[...]

> As long as there is an infinite common set, the intersection is this infinite common set.

Another epiphany! Wow!

Unfortunately, if you have an infinite collection of end segments, then no end segment E(n+1) can contain any natural number contained in FISON(n). Since an infinite collection of end segments does not have an element with maximum index K, this means that the intersection is empty.

[William]

Nope. The "infinite common set" changes. The intersection does not change

[FromTheRafters]

Fritz Feldhase explained on 9/4/2022 :

There is no 'FISON' ending in zero, so no 'endsegment' starting at one.

[Fritz Feldhase]

WHAT'S THE MATTER WITH YOU FUCKING ASSHOLE FULL OF SHIT?!

Hint:

FISONs: F(1) = {1}, F(2) = {1, 2}, F(3) = {1, 2, 3}, ...
endsegments: E(1) = {1, 2, 3, ...}, E(2) = {2, 3, 4, ...}, E(3) = {3, 4, 5, ...}

AND NOW FUCK OF, YOU SILLY ASSHOLE!

[Fritz Feldhase]

On Sunday, September 4, 2022 at 3:25:38 PM UTC+2, FromTheRafters wrote:
> Fritz Feldhase explained :
> >
> > Proof (by induction): E(1) = IN, hence E(1) is infinite. [...]

> >
> E(1) does not equal the naturals in muckymath.

*sigh*

Hint: "With E(1) = ℕ all is said. But that should be known meanwhile." (WM, sci.math)

Hope this helps.

[Sergio]

so F(0) and E(0) are up for grabs ?



I hereby, declare and now royally assign forever: F(0) = {} AND E(0) as the set of counting numbers + 0 = {0, 1, 2, 3, ...}

EVERYBODY GOT THAT ?

For completeness, What about F(-k) and E(-k) ?? they apply in the case where you move sheeps from one field to another field, lose count and have to
start over.

[Sergio]

remember muckymath(t) is time dependent

muckymath(t1) =/= muckymath(t2)

[Fritz Feldhase]

On Sunday, September 4, 2022 at 6:52:32 PM UTC+2, Sergio wrote:
> On 9/4/2022 11:19 AM, Fritz Feldhase wrote:

> > On Sunday, September 4, 2022 at 6:02:19 PM UTC+2, FromTheRafters wrote nonsense.

> >
> > WHAT'S THE MATTER WITH YOU FUCKING ASSHOLE FULL OF SHIT?!
> >
> > Hint:
> >
> > FISONs: F(1) = {1}, F(2) = {1, 2}, F(3) = {1, 2, 3}, ...
> > endsegments: E(1) = {1, 2, 3, ...}, E(2) = {2, 3, 4, ...}, E(3) = {3, 4, 5, ...}
> >
> > AND NOW FUCK OF, YOU SILLY ASSHOLE!
> >
> so F(0) and E(0) are up for grabs ?

Note that 0 !e IN in this context (i. e. for WM).

Now we have:

F: IN --> P(IN) with F(n) = {m e IN : m <= n} for all n e IN.
E: IN --> P(IN) with E(n) = {m e IN : m >= n} for all n e IN.

Hence "F(0)" and "E(0)" are not defined.

[FromTheRafters]

Fritz Feldhase presented the following explanation :

You are confused. He takes the elements from the set of natural numbers
and partitions them into two complementary ordered sets.

{1,2,3,4,5} <---- Initial Segment of Naturals. Natural numbers less
than or equal to n -- in this case n equals five.

https://proofwiki.org/wiki/Definition:Initial_Segment_of_Natural_Numbers

{6,7,8,9,...} <---- and its complement, the endsegment, whose index is,
in this case, n+1 or six.

the index for E(1) would have to be F(0)+1 except WM has no zero as an
element of the naturals.

[Fritz Feldhase]

I said FUCK OF, YOU FUCKING ASSHOLE FULL OF SHIT!

[FromTheRafters]

It happens that Sergio formulated :

F(0) is all natural numbers less than or equal to zero. For muckymath,
where zero is not a natural number (when he doesn't want it to be) it
doesn't exist. The associated endsegment starts at undefined plus one.

> I hereby, declare and now royally assign forever: F(0) = {} AND E(0) as the
> set of counting numbers + 0 = {0, 1, 2, 3, ...}

Here the index of the endsegment should match the cardinality of the
FISON which is n+1 so F(0) should be associated with E(1) shouldn't it?

> EVERYBODY GOT THAT ?

In that case, a FISON is all natural numbers strictly less than n which
makes n the index for the endsegment, and n minus one the 'greatest'
element in the FISON.

> For completeness, What about F(-k) and E(-k) ?? they apply in the case where
> you move sheeps from one field to another field, lose count and have to start
> over.

Mitch would be so ashamed of you for even suggesting such a thing.

[FromTheRafters]

Calm down, and off is spelled with two Fs.

[FromTheRafters]

on 9/4/2022, Fritz Feldhase supposed :

It doesn't. The index for the endsegment should be the same as the
greatest element of the associated FISON *plus* one. Where is the
associated FISON for E(1) if E(1) is all of the natural numbers?

[Ross A. Finlayson]

"Vector methods: 1) addition of vectors, ..."

[Ross A. Finlayson]

x + -x

[Ross A. Finlayson]

Oh, this is where instead "it's all the way to the end and back".

"The counting forward is same as counting backward: except one is
learning the count and the other is knowing the count".

[Fritz Feldhase]

Right.

So FUCK OFF, YOU FUCKING ASSHOLE FULL OF SHIT!

[Fritz Feldhase]

On Sunday, September 4, 2022 at 9:07:26 PM UTC+2, FromTheRafters wrote:
> on 9/4/2022, Fritz Feldhase supposed :
> > On Sunday, September 4, 2022 at 3:25:38 PM UTC+2, FromTheRafters wrote:
> >> Fritz Feldhase explained :
> >>>
> >>> Proof (by induction): E(1) = IN, hence E(1) is infinite. [...]
> >>>
> >> E(1) does not equal the naturals in muckymath.
> >
> > *sigh*
> >
> > Hint: "With E(1) = ℕ all is said. But that should be known meanwhile." (WM,
> > sci.math)
> >
> > Hope this helps.
> >
> It doesn't.

"There was no hope for him this time: it was the third stroke." (James Joyce, Dubliners)

[FromTheRafters]

Fritz Feldhase submitted this idea :

That's better.

[mitchr...@gmail.com]

On Thursday, September 1, 2022 at 9:24:43 AM UTC-7, William wrote:
> On Thursday, September 1, 2022 at 10:09:42 AM UTC-3, WM wrote:
> >
> > If you argue that all numbers leave the intersection but [a set of] infinitely many numbers do not leave the contributing endsegments, then you are outside of logic and reason.
>
> Indeed. But there is no unchanging set of infinitely many numbers that does not "leave the endsegments" . There is only a decreasing sequence of sets
> with no last element.

Who is going to put together infinite element set? A function projected by man still doesn't ever complete infinity.

[zelos...@gmail.com]

torsdag 1 september 2022 kl. 21:20:17 UTC+2 skrev mitchr...@gmail.com:
> On Thursday, September 1, 2022 at 9:24:43 AM UTC-7, William wrote:
> > On Thursday, September 1, 2022 at 10:09:42 AM UTC-3, WM wrote:
> > >
> > > If you argue that all numbers leave the intersection but [a set of] infinitely many numbers do not leave the contributing endsegments, then you are outside of logic and reason.
> >
> > Indeed. But there is no unchanging set of infinitely many numbers that does not "leave the endsegments" . There is only a decreasing sequence of sets
> > with no last element.

> How do you contain infinite set elements?
>
> Mitchell Raemsch

A set either has infinite or finite cardinality. There is no "how", it just does. Because it is not physical

[zelos...@gmail.com]

torsdag 1 september 2022 kl. 21:51:06 UTC+2 skrev mitchr...@gmail.com:
> On Thursday, September 1, 2022 at 12:28:31 PM UTC-7, FromTheRafters wrote:
> > mitchr...@gmail.com explained :

> > > On Thursday, September 1, 2022 at 9:24:43 AM UTC-7, William wrote:
> > >> On Thursday, September 1, 2022 at 10:09:42 AM UTC-3, WM wrote:
> > >>>
> > >>> If you argue that all numbers leave the intersection but [a set of]
> > >>> infinitely many numbers do not leave the contributing endsegments, then you
> > >>> are outside of logic and reason.
> > >>
> > >> Indeed. But there is no unchanging set of infinitely many numbers that does
> > >> not "leave the endsegments" . There is only a decreasing sequence of sets
> > >> with no last element.
> > >
> > > How do you contain infinite set elements?

> > With curly brackets.
> That can't be done. Just as there are no countable infinities.
of course there is, N is countably infinite, so is Q, Z and many others, R is uncountably infinite.

[zelos...@gmail.com]

It is not a computer function you retard

the set N is already infinite.

[WM]

Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:36:31 UTC+2:
> On Sunday, 4 September 2022 at 11:08:34 UTC-3, WM wrote:
> [...]
> > As long as there is an infinite common set, the intersection is this infinite common set.
>

> Unfortunately, if you have an infinite collection of end segments, then no end segment E(n+1) can contain any natural number contained in FISON(n).

An infinite endsegment has an infinite contents together with all infinite endsegments.

> Since an infinite collection of end segments does not have an element with maximum index K, this means that the intersection is

dark.

Regards, WM

[WM]

The intersection of infinite endsegments is infinite.

Regards, WM

[WM]

FromTheRafters schrieb am Sonntag, 4. September 2022 um 21:07:26 UTC+2:
> on 9/4/2022, Fritz Feldhase supposed :
> > On Sunday, September 4, 2022 at 3:25:38 PM UTC+2, FromTheRafters wrote:
> >> Fritz Feldhase explained :
> >>>
> >>> Proof (by induction): E(1) = IN, hence E(1) is infinite. [...]
> >>>
> >> E(1) does not equal the naturals in muckymath.
> >
> > *sigh*
> >
> > Hint: "With E(1) = ℕ all is said. But that should be known meanwhile." (WM,
> > sci.math)
> >
> > Hope this helps.
> It doesn't. The index for the endsegment should be the same as the
> greatest element of the associated FISON *plus* one.

That would be possible, perhaps even better. But I have originally defined it in this way:
F(n) = {1, 2, 3, ..., n}
E(n) = {n, n+1, n+2, ...}
E(1) = ℕ = {1, 2, 3, ...}
And it would cause confusion to change that.

Regards, WM

[FromTheRafters]

WM has brought this to us :

E(n) then intersects F(n) at n when they are supposed to be the
disjoint (complementary) subsets of the newly partitioned naturals. But
okay, muckymath is bulletproof with the axiom of because I (WM) said
so.

Carry on.

[zelos...@gmail.com]

It isn't, you assert it but it isn't!

[Sergio]

On 9/6/2022 5:46 AM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:36:31 UTC+2:
>> On Sunday, 4 September 2022 at 11:08:34 UTC-3, WM wrote:
>> [...]
>>> As long as there is an infinite common set, the intersection is this infinite common set.
>>
>> Unfortunately, if you have an infinite collection of end segments, then no end segment E(n+1) can contain any natural number contained in FISON(n).
>
> An infinite endsegment has an infinite contents together with all infinite endsegments.

Wrong

corrected: An endsegment has an infinite number of elements.

Fact: The intersection of all endsegments is empty.

>
>> Since an infinite collection of end segments does not have an element with maximum index K, this means that the intersection is
>

> .

Empty.

>
> Regards, WM

[Sergio]

Corrected: The intersection of all endsegments is empty.


>
> Regards, WM

Troll.

[Ross A. Finlayson]

Think of it this way: "it had carried on then for whatever reason,
the condition changed: it's the same axiomatic system, including a dynamic".

That simply changes what is the parallel condition in the finite,
to be just the same way, axiomatic in dynamics.

Then, all your moot arguments are compressible to a pinhead.

[Sergio]

On 9/6/2022 5:55 AM, WM wrote:
> FromTheRafters schrieb am Sonntag, 4. September 2022 um 21:07:26 UTC+2:
>> on 9/4/2022, Fritz Feldhase supposed :
>>> On Sunday, September 4, 2022 at 3:25:38 PM UTC+2, FromTheRafters wrote:
>>>> Fritz Feldhase explained :
>>>>>
>>>>> Proof (by induction): E(1) = IN, hence E(1) is infinite. [...]
>>>>>
>>>> E(1) does not equal the naturals in muckymath.
>>>
>>> *sigh*
>>>
>>> Hint: "With E(1) = ℕ all is said. But that should be known meanwhile." (WM,
>>> sci.math)
>>>
>>> Hope this helps.
>> It doesn't. The index for the endsegment should be the same as the
>> greatest element of the associated FISON *plus* one.
>
> That would be possible, perhaps even better. But I have originally defined it in this way:

You did not define it.

[Ross A. Finlayson]

It's only copies of references of real mathematical resources, anyways.

[Sergio]

so if he stops at k (FISON), then he really already started at k (Endsegment) ?

so the intersection of FISON(k) and Endesgment(k) is k

[Ross A. Finlayson]

In his McDuck cart with you right along, ....

[Ross A. Finlayson]

The axiomatic and dynamic, is much simpler and better,
than argument, in terms of unending argument and regress.

Though it involves reasoning facilities, axiomatics is dynamic
in the sense of the very strong axiomatic, what must entail
all reasoning facilities: at least this simple axiomatic is simply
dynamic: whatever all faults in reasoning reduce to induction.

The utterly and fundamentally technical, here for the infinite
there's that WM's failed induction over a related rates system,
because his relation that is proportional is first not second
that there is an infinite rate, (respectively not a largest finite rate).

Then, having the proportional that runs out together is a related
rate, any matter of proportion in time.

This assumes a linear constant rate of time at least from every
object in space-time, or the local rate down to zero.

Which leaves though both sides having brought down the
others' reasoning, while still having to leave them together.

Then, this leads immediately to my slates and the state of
modern axiomatics in continuum analysis, neatly out.

[Ross A. Finlayson]

"It's easy to be a troll when mathematics does all the work of being right. "


"Shannon is sometimes given credit for Nyquist theorem,
where it may be so that Shannon and "continuity is result
of rationals" have that Shannon's results in signal theory
and for information theory, for information theory, and,
signal theory, with respect to formalism and here with
formalism in foundations, have that "no, it is not so that
'continuity is result of rationals', it's not to be implied
that it's a point of Shannon's when connecting signal theory
and properties of continuous sampling theorems, except
to illustrate the entire connection here in the abstract."

Then, that "I said so", is basically that mostly I have
studied the formalism of the line continuity and the
field continuity, with respect to the signal continuity,
and continuous functions which is the usual formalism. "

[Gus Gassmann]

On Tuesday, 6 September 2022 at 07:46:23 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:36:31 UTC+2:
> > On Sunday, 4 September 2022 at 11:08:34 UTC-3, WM wrote:
> > [...]
> > > As long as there is an infinite common set, the intersection is this infinite common set.
> >
> > Unfortunately, if you have an infinite collection of end segments, then no end segment E(n+1) can contain any natural number contained in FISON(n).
> An infinite endsegment has an infinite contents together with all infinite endsegments.

This gobbledygook of yours is so grotesque, it is not even false.

> > Since an infinite collection of end segments does not have an element with maximum index K, this means that the intersection is
> dark.

Well, you seem to be taking "dark" and "empty" as synonymous. Very well. Your skull cavity is both empty and dark.

[Sergio]

WM denies axioms exist, however he uses his statements as axioms

>
> Though it involves reasoning facilities, axiomatics is dynamic
> in the sense of the very strong axiomatic, what must entail
> all reasoning facilities: at least this simple axiomatic is simply
> dynamic: whatever all faults in reasoning reduce to induction.
>
> The utterly and fundamentally technical, here for the infinite
> there's that WM's failed induction over a related rates system,
> because his relation that is proportional is first not second
> that there is an infinite rate, (respectively not a largest finite rate).

yea, today's infinite rate is the highest in 40 years!

[Sergio]

agree, its abstraction.

>
> Then, that "I said so", is basically that mostly I have
> studied the formalism of the line continuity and the
> field continuity, with respect to the signal continuity,
> and continuous functions which is the usual formalism. "

line integrals too ?

[Ross A. Finlayson]

Sure, mathematics does all the work of being right.

Line integrals are difficult also rudimentary.

Solving all WM's problems in related rates,
and with assigning the limits as opaque,
results why what falls out.

The line integral, introduced after the area integral,
and for surface integrals in a similar sense, and
the path of integration, is that there is the path integral.

Then, it is also some "bridge integral". I.e., the path
is defined according to its bounds, what for example, is
the "path integral, add up where it's been", for the
bridge integral, partition where it's going".

The line integral is often contrived "end-of-the-line" interal,
i.e., "a line integral is always attached". But, of course
it's still the exact same interpretation, "the length of the line".


Now, I just introduced or "defined", bridge integral, though
it's usually just the shortest line integral, that also happens
for attachments across the square or the angle, what's drawn out.

(Square, radial, anti-logarithmic, ....)

That the differential is anti-integral is vice-versa.

The antilogarithm or power, is for tables: under linear inputs, usually.

Paint cans are volume but usually define what area they cover.

Why are you bringing line integral into it? Isn't that contrived
for either the catenary or parabolic?

[Ross A. Finlayson]

On Saturday, September 3, 2022 at 10:08:04 AM UTC-7, Ross A. Finlayson wrote:
> On Friday, September 2, 2022 at 7:17:37 PM UTC-7, Sergio wrote:
> > On 9/2/2022 3:13 PM, WM wrote:
> > > Fritz Feldhase schrieb am Freitag, 2. September 2022 um 16:42:51 UTC+2:
> > >> On Friday, September 2, 2022 at 2:39:00 PM UTC+2, WM wrote:
> > >>> William schrieb am Donnerstag, 1. September 2022 um 22:49:27 UTC+2:
> > >>>>
> > >>>> b_n = [n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n).
> > >>
> > >> Well, let's say IN n [n,ω) is an interval of natural numbers. (ω is not an element of b_n for any n).
> > >>
> > >>> The intersection is empty
> > >>
> > >> Right.
> > >>
> > >>> No,
> > >>
> > >> Yes.
> > >>
> > >>> the intersection is infinite
> > >>
> > >> No, it isn't.

> > >
> > > As soon as there is no longer an infinite common set,

> > there is no common infinite set.
> >
> >
> > Go back to your potentially infinite book, and find none of the math there applies to infinite sets.
> > > the next state of affairs is a finite set by the very definition of endsegments, i.e., simple mathematics: ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> > that equation does not define an endsegment.
> > >
> > > To overlook this shows incompetence in mathematics and logic.
> > which you have in spades.
> > >
> > > To deny it by the analogy that
> > > ∀k ∈ ℕ ∃m ∈ ℕ: k < m ==> ∃m ∈ ℕ ∀k ∈ ℕ: k < m
> > > is an invalid conclusion, is an invalid conclusion and shows naivety.
> > red herring
> >
> > >
> > > Regards, WM
> No it's:
>
> ∀k ∈ ℕ ∃m ∈ ℕ: k < m ==> ∀m2 ∈ ℕ ∃k2 ∈ ℕ ∃m3 ∈ ℕ: k2 < m3, ∃j ∈ ℕ: k2 < m, ~∃l ∈ ℕ, l < j, ~(m2< m3 ^ ~ m2 < m3)
>
>
> I.e. there is "always only proving the properties of natural numbers
> that have exists m>n while also not for zero".
>
> I.e., saying "exists m , ..., in N", is follows "exists N, exists < in N, exists m in N, ...".

That it's "<->".

Writing out what combines mutual implication instead of some otherwise
plain direct implication, is besides cases in logic, conclusions in logic.

[FromTheRafters]

on 9/6/2022, Sergio supposed :

It looks that way to me. His first FISON's end should be one less than
his index for its associated endsegment.

These are well ordered sets and E(1) implies an associated ISON ending
in zero -- which is undefined in |N*. His

You can't just have a successor function, you need an initial element.
Zero is a bad choice for an initial element in the non-zero naturals.

There are those whom take the entire set {0,1,2,3,...} to be an
endsegment, but that was not stipulated in his "definition" and can
create more problems with his wishy-washy axioms.

[Fritz Feldhase]

On Tuesday, September 6, 2022 at 4:12:51 PM UTC+2, Sergio wrote:

> so the intersection of FISON(k) and Endesgment(k) is k

No. It's {k}.

[Fritz Feldhase]

On Tuesday, September 6, 2022 at 6:12:07 PM UTC+2, FromTheRafters wrote:
> on 9/6/2022, Sergio supposed :

> >
> > so the intersection of FISON(k) and Endesgment(k) is k
> >
> It looks that way to me.

Then look again, since it is {k}, not k.

[Gus Gassmann]

It's {k}not easy... SCNR

[WM]

zelos...@gmail.com schrieb am Dienstag, 6. September 2022 um 14:38:11 UTC+2:

> > The intersection of infinite endsegments is infinite.
> >

> It isn't, you assert it but it isn't!

Then all endsegments contain infinite sets, but no element of one endsegment E(k) is in the infinite sets of some other emdsegments. What are they consisting of?

Regards, WM

[Sergio]

On 9/7/2022 8:11 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 6. September 2022 um 14:38:11 UTC+2:
>
>>> The intersection of infinite endsegments is infinite.
>>>
>> It isn't, you assert it but it isn't!
>
> Then all endsegments contain infinite sets,

Wrong.
1 All endsegments are infinite by definition.
2 All endsegments contain elements, not sets.

> but no element of one endsegment E(k) is in the infinite sets of some other emdsegments.

Wrong. 3 is in E(1), E(2), E(3)

> What are they consisting of?

You tell me, its your imagination posting this garbage.

>
> Regards, WM

[Gus Gassmann]

Boring!!!! Every end segment E(n) is the set of natural numbers {n + k: k = 0 OR k in |N}. And yes, it is clear that for every n in |N, card(E(n)) = card(|N). Your continued babbling to the contrary is just that --- babbling, without any substance, proof, or even motivation.

[Sergio]

google line integrals,
Just testing to see how much math you know, like a sounding, this one is by the Deep One less four quarters.

[Sergio]

correctomund{o}! it is a set for sure.

[Fritz Feldhase]

On Wednesday, September 7, 2022 at 3:31:24 PM UTC+2, Gus Gassmann wrote:
> On Wednesday, 7 September 2022 at 10:11:28 UTC-3, WM wrote:
> >

> > [...] but no element of one endsegment E(k) is in the infinite sets of some other endsegments. What are they consisting of?

What does he even want claim here? Looks like some sort of a quantifier shift to me.

[Ross A. Finlayson]

Come on now, Fritz, "Come on, WM, even Ross tries to save you".


The quantifier shift at all, indeed.

Universal quantifier? Make four:
for-each, for-any, for-every, for-all.
There are orders implied by the quantifiers,
and orders implied in the quantifiers.

Existential quantifier? Unique existential quantifier.

[Ross A. Finlayson]

"That was the greatest comic tour de force that anybody had seen in a long time."

(Roger Rabbit)


The line integral was introduced after the catenary problem,
basically at the end of otherwise a chapter in the text, instead
of starting a new chaper, the "advanced calculus". So, I wanted
to make the line integral the same as the quarter-bend, but instead
it was both amounting to the catenary via parabola, and, that
I want to solve the catenary in the parabola, but, otherwise the
line integral was let down to line elements and ds.

Then the surface integral, after surfaces of revolution, or paint cans,
then the path of integration is free for what it is, what it amounts to.

So, I'd imagine several line integral setups for usual problems of
what is the length according to perimeter, circumference, besides
as wise the square, and quadratic, in the elements, and the paths.
This is where according to the paths there are elements that are
integrable, that any contrivance in symmetry or exhaust, or relation
in measure here length, is for usual geometric and stateful examples,
in terms of how and why they are line elements and path elements.

The "advanced calculus", ..., here when it got to "yes I have derived this
Green's function accordingly, but, what I want is instead for example
different elements", then though later after gradient descent and
Stokes, then again out through the Hamiltonian what really though
is for the Hermitian, then to flow and what is fluid mechanics and
fluid dynamics, then again it was "still these are Green's functions
after Stokes, 'advanced calculus'", I've studied the advanced calculus,
and have several texts, but surely most my exposure is peripheral.

I.e. the brief tableau for the determinantal, the "dot product is ...
derivative anti-integral placeholder-element", is for people to
learn inner and outer products and wedge and dot and so on.

And why and how it's computable.

Read this "Exterior derivative" Wiki: https://en.wikipedia.org/wiki/Exterior_derivative
Then you'll notice it's as if I intend to paste the above into it.

[Ross A. Finlayson]

https://en.wikipedia.org/wiki/Category:Generalized_functions

[Sergio]

https://en.wikipedia.org/wiki/Pseudo-differential_operator

WM is a Pseudo-math operator

[William]

On Tuesday, September 6, 2022 at 7:49:45 AM UTC-3, WM wrote:
> William schrieb am Sonntag, 4. September 2022 um 16:44:45 UTC+2:
> > On Sunday, September 4, 2022 at 11:08:34 AM UTC-3, WM wrote:
> > > William schrieb am Samstag, 3. September 2022 um 22:29:06 UTC+2:

> > > > On Saturday, September 3, 2022 at 5:12:27 PM UTC-3, WM wrote:
> >
> > > > > As soon as there is no longer an infinite common set,

> > > > This does not happen.

> > > As long as there is an infinite common set, the intersection is this infinite common set.

> > Nope. The "infinite common set" changes. The intersection does not change

> The intersection of infinite endsegments is infinite.

Nope. The intersection of a set of endsegments with smallest endsegment is infinite. The intersection of a set of endsegments with no smallest endsedgment is empty.

[WM]

William schrieb am Mittwoch, 7. September 2022 um 19:20:35 UTC+2:
> On Tuesday, September 6, 2022 at 7:49:45 AM UTC-3, WM wrote:

> > > > As long as there is an infinite common set, the intersection is this infinite common set.
> > > Nope. The "infinite common set" changes.

But it remains infinite.

> The intersection does not change

Only if there is a last endsegment.

> > The intersection of infinite endsegments is infinite.

> Nope.

Yes. Here is the proof:
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
Find an endsegment that is not addressed here and indexed by an elements of the infinite set ℕ.

> The intersection of a set of endsegments with smallest endsegment is infinite.

Yes.

> The intersection of a set of endsegments with no smallest endsedgment is empty.

No, it is not existing.

Only the intersection of a set of endsegments with empty endsegment is empty. Inclusion monotony.

Regards, WM

[mitchr...@gmail.com]

How are you going to put together an infinite set of infinite elements?
You can't assemble infinity complete outside of the Continuum Hypothesis.


Mitchell Raemsch

[FromTheRafters]

WM formulated on Wednesday :

> zelos...@gmail.com schrieb am Dienstag, 6. September 2022 um 14:38:11 UTC+2:
>
>>> The intersection of infinite endsegments is infinite.
>>>
>> It isn't, you assert it but it isn't!
>
> Then all endsegments contain infinite sets

Yes, if they are defined as FISON complement sets. With FISON {1,2,3,4}
we automatically have endsegment {5,6,7,8,...} which contains, for
example, {5,7,9,11,...} and {6,8,10,12,...} as subsets which are both
infinite sets.

What good does this do you in your quest?